How do i solve the integral $ \int_{-\infty}^{\infty} x^3*e^{-ax^2}dx $ I have a hard time doing this problem can anyone help me please ?
 A: Using integration by parts once gives
$$\int x^3e^{-ax^2}\mathrm{d}x=-\frac{x^2}{2a}e^{-ax^2}+\frac1a\int xe^{-ax^2}dx=-\frac{x^2}{2a}e^{-ax^2}-\frac1{2a^2}e^{-ax^2}+C$$
Hence the intgeral is equal to
$$\begin{align}
\int_{-\infty}^\infty x^3e^{-ax^2}\mathrm{d}x
&=\lim_{b\to\infty}\int_{-b}^0 x^3e^{-ax^2}\mathrm{d}x+\lim_{c\to\infty}\int_0^c x^3e^{-ax^2}\mathrm{d}x\\
&=\lim_{b\to\infty}\left[-\frac{x^2}{2a}e^{-ax^2}-\frac1{2a^2}e^{-ax^2}\right]_{-b}^0+\lim_{c\to\infty}\left[-\frac{x^2}{2a}e^{-ax^2}-\frac1{2a^2}e^{-ax^2}\right]_0^c\\
&=\lim_{b\to\infty}\left(-\frac1{2a^2}+\frac{b^2}{2a}e^{-ab^2}+\frac1{2a^2}e^{-ab^2}\right)+\lim_{c\to\infty}\left(-\frac{c^2}{2a}e^{-ac^2}-\frac1{2a^2}e^{-ac^2}+\frac1{2a^2}\right)\\
&=-\frac1{2a^2}+\frac1{2a^2}\\
&=0\\
\end{align}$$
A: let $f(x)=x^3e^{-ax^2}$
Since the integral is in the form
$$\int_{-b}^{b} f(x) dx$$
If we can show the function is odd then the area is 0 because it cancels out.
$$-f(x)=f(-x)$$
$$-x^3e^{-ax^2}=(-x)^3e^{-a(-x)^2}$$
$$-x^3e^{-ax^2}=-x^3e^{-ax^2}$$
Hence the answer is $0$.
A: Noticing that the integrand is an odd function allows us to show that the integral converges (by comparison test, for example), then conclude that the integral is $0.$ However, we can get there even if we don't notice that.
First of all, note that $$\frac{d}{dx}\left[e^{-ax^2}\right]=-axe^{-ax^2}$$ by Chain Rule, regardless of $a.$
Now, if $a=0,$ then the integral doesn't converge, so I assume that $a$ is non-$0,$ and that you just didn't mention it. By my assumption, we can conclude that $$xe^{-ax^2}=\frac{d}{dx}\left[-\frac1ae^{-ax^2}\right],$$ which will allow us to use integration by parts to determine an antiderivative.
\begin{eqnarray}\int x^3e^{-ax^2}dx &=& \int x^2\cdot\frac{d}{dx}\left[-\frac1ae^{-ax^2}\right]dx\\ &=& x^2\cdot-\frac1ae^{-ax^2}-\int\frac{d}{dx}\left[x^2\right]\cdot-\frac1ae^{-ax^2}dx\\ &=& -\frac1ax^2e^{-ax^2}+\frac2a\int xe^{-ax^2}dx\\ &=& -\frac1ax^2e^{-ax^2}+\frac2a\int \frac{d}{dx}\left[-\frac1ae^{-ax^2}\right]dx\\ &=& -\frac1ax^2e^{-ax^2}+\frac2a\cdot-\frac1ae^{-ax^2}\\ &=& -\frac1ax^2e^{-ax^2}-\frac2{a^2}e^{-ax^2}\\ &=& -\left(\frac1ax^2+\frac2{a^2}\right)e^{-ax^2}\end{eqnarray}
Thus,
\begin{eqnarray}\int_{-\infty}^\infty x^3e^{-ax^2}dx &=& \lim_{b\to-\infty,c\to\infty}\int_b^c x^3e^{-ax^2}dx\\ &=& \lim_{b\to-\infty,c\to\infty}-\left(\frac1ac^2+\frac2{a^2}\right)e^{-ac^2}+\left(\frac1ab^2+\frac2{a^2}\right)e^{-ab^2}\\ &=& -\lim_{c\to\infty}\left(\frac1ac^2+\frac2{a^2}\right)e^{-ac^2}+\lim_{b\to-\infty}\left(\frac1ab^2+\frac2{a^2}\right)e^{-ab^2}\\ &=& -\lim_{c\to\infty}\left(\frac1ac^2+\frac2{a^2}\right)e^{-ac^2}+\lim_{c\to\infty}\left(\frac1a(-c)^2+\frac2{a^2}\right)e^{-a(-c)^2}\\ &=& -\lim_{c\to\infty}\left(\frac1ac^2+\frac2{a^2}\right)e^{-ac^2}+\lim_{c\to\infty}\left(\frac1ac^2+\frac2{a^2}\right)e^{-ac^2}.\end{eqnarray} Consequently, if you can show that $\lim_{c\to\infty}\left(\frac1ac^2+\frac2{a^2}\right)e^{-ac^2}$ exists, then you'll have proved that the integral is $0.$
